KPK 9th Class Physics Chapter 6 Work and Energy Short Questions Answers
KPK 9th Class Physics Chapter 6 Work and Energy Short Questions with answers are combined for all 9th class(Matric/ssc) Level students.Here You can prepare all Physics Chapter 6 Work and Energy short question in unique way and also attempt quiz related to this chapter.Just Click on Short Question and below Answer automatically shown. After each question you can give like/dislike to tell other students how its useful for each.
Class/Subject: 9th Class Physics
Chapter Name: Work and Energy
Board: All KPK Boards
- Malakand Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Mardan Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Peshawar Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Swat Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Dera Ismail Khan Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Kohat Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Abbottabad Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
- Bannu Board 9th Class Physics Chapter 6 Work and Energy short questions Answer
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KPK 9th Class Physics Chapter 6 Work and Energy Short Questions Answers
No, centripetal force never do any work on an object. The centripetal force compels a body to move in a circle. The centripetal force is always perpendicular to the direction of the motion. i.e 0 = 90°
And we know that:
Work = FS cos0
= FScos90° cos90°
Work = 0
To do any work the components of the force must be in the direction of motion. The centripetal force has no such components, so centripetal force never do work.
K.E of the bullet of mass m1 will be:
(K.E)1=1/2m1v12 …………(1)
Similarly the K.E of bullet of mass ‘m2‘ will be?”:
(K.E)2=1/2m2v22 ………… (2)
Since both the bullets have same K.E but the mass of one bullet is twice of the other bullet (i.e m1=2m2).
Now,
(K.E)1= (K.E)2
1/2m1v12=1/2m2v22
1/2(2m2)v12=1/2m2v22 (∴m1=2m2)
2m2v12=m2v22
2m2v12/m2= v22
v22= 2v12
Talking square root on both sides:
√v22= √2√v12
v2= √2v1
Thus the bullet of mass “m2“ which is lighter than “m1” travels with great speed with a facts of √2.
The bullet of mass “m2” covers a greater distance than “m1” so bullet having mass “m2” do the most work.
P.E= mgh
P.E depends upon the height from the surface of earth. If the object remains at the same elevational from the surface of earth the gravitational P.E will remain same. Gravitational P.E is increased with the increase of height.
1. When the road are wind up, due to the small angle of inclination we can do less work against the gravity instead of straight up situation.
2. By increasing the frictional force between road and tyres the road leading to the top of a mountain wind back and forth.
3. To avoid the skidding force the road leading to top a mountain wind back and forth.
Since
K.E=1/2 mv2
If we double the mass only then m’ =2m we get.
K.E‘=1/2×2m ×v2
K.E=2 (1/2mv2)
(K.E) = 2K.E
If we double the velocity only then:
V “ = 2v
(K.E)” = 1/2 m (2v)2
(K.E)” = 1/2×m×4v2
(K.E)” = 4(1/2mv2)
(K.E)” = 4 K.E
This result shows that greater effect on the kinetic energy is produced by doubling the velocity.
K.E = 1/2mv2……………(1)
If the speed of the particle is tripled then V’ = 3V.
Now equation (1) can be written as:
(K.E)’ =1/2mv2
Put v’ = 3v
K.E)’ = 1/2m (3v)2
= 1/2m9.v2
= 9(1/2mmv2)
(K.E)’ = 9K.E (∴K.E=1/2mv2)
When the speed of the particle increase to triple then K.E increases 9 times.
The rate of doing work is called power.
Power = work/time
Here work = P.E =mgh
Power = mgh/t…………..(1)
When height of beam because double in half time, then above equation can be written as:
P’ = mgh‘/t‘ …………………(2)
Where h’ = 2h
And t’=t/2
Now equ: (2) becomes:
P’ = mg×2h/t
P’ = 2mg×2/t
P’ = 4mgh/t
P’=4P
∴P= mgh/t
The motor’s power increases by four times.
